Saline solution may be needed for a variety of purposes; for example, it is part of some traditional medicine. So how to prepare a 1% solution if you don’t have special beakers at home to measure the amount of product? In general, even without them you can make a 1% salt solution. How to prepare it is described in detail below. Before you start preparing such a solution, you should carefully study the recipe and determine exactly the necessary ingredients. The thing is that the definition of “salt” can refer to different substances. Sometimes it turns out to be ordinary table salt, sometimes rock salt, or even sodium chloride. As a rule, in a detailed recipe it is always possible to find an explanation of which particular substance is recommended to be used. In folk recipes, magnesium sulfate is also often indicated, which has the second name “Epsom salt”.
If the substance is required, for example, to gargle or relieve pain from a tooth, then most often in this case it is recommended to use a saline solution of sodium chloride. In order for the resulting product to have healing properties and not cause harm to the human body, only high-quality ingredients should be selected for it. For example, rock salt contains a lot of unnecessary impurities, so it is better to use regular fine salt instead (iodized salt can also be used for rinsing). As for water, at home you should use filtered or at least boiled water. Some recipes recommend using rainwater or snow. But, given the current environmental state, this is not worth doing. Especially for residents of large cities. It's better to just thoroughly clean the tap water.
If you don’t have a special filter at home, you can use the well-known “old-fashioned” method to purify water. It involves freezing tap water in the freezer. As you know, in the process, it is the purest liquid that first turns into ice, and all harmful impurities and dirt sink to the bottom of the container. Without waiting for the entire glass to freeze, you should remove the top ice part and then melt it. Such water will be as clean and safe for health as possible. It is this that can be used to prepare a saline solution.
Now it’s time to decide on the units of measurement for liquids and solids. For salt, it is most convenient to use a teaspoon. As you know, it holds 7 grams of product, if the spoon is heaped, then 10. The latter option is more convenient to use for calculating the percentage. It’s easy to measure water with an ordinary cut glass if you don’t have special beakers at home. It contains 250 milliliters of water. The mass of 250 milliliters of pure fresh water is equal to 250 grams. It is most convenient to use half a glass of liquid or 100 grams. Next is the most difficult stage of preparing the saline solution. It’s worth once again carefully studying the recipe and deciding on the proportions. If it is recommended to take a 1% salt solution, then in every 100 grams of liquid you will need to dissolve 1 gram of solid. The most accurate calculations will tell you that you will need to take 99 grams of water and 1 gram of salt, but such precision is unlikely to be required.
It is quite possible to make some error and, for example, add one heaped teaspoon of salt to one liter of water to obtain a 1% saline solution. Currently, it is often used, for example, in the treatment of colds and especially sore throat. You can also add soda or a few drops of iodine to the finished solution. The resulting gargle mixture will be an excellent effective and efficient remedy for a sore throat. The discomfort will go away after just a few procedures. By the way, such a solution is not prohibited for use by the youngest members of the family. The main thing is not to overdo it with additional ingredients (especially iodine), otherwise you can damage the mucous membrane of the oral cavity and only aggravate the condition of a sore throat.
Also, a saline solution can be used to relieve a nagging, aching toothache. True, it is more effective to use a more saturated one, for example, 10 percent. This mixture can really relieve painful discomfort in the oral cavity for a short time. But it is not a medicine, so under no circumstances should you put off visiting the dentist after relief.
Alkali solutions. Caustic alkalis and their solutions actively absorb moisture and carbon dioxide from the air, so preparing solutions of precise titre from them is difficult. It is best to make such solutions from fixanals. To do this, take a test tube with a fixation of the required normality and a 1 liter volumetric flask. A glass funnel with a glass striker inserted into it, the sharp end of which faces upward, is inserted into the flask.
When the striker is properly positioned in the funnel, the ampoule containing the fixanal is allowed to fall freely so that the thin bottom of the ampoule breaks when it hits the sharp end of the striker. After this, the side recess of the ampoule is punched and the contents are allowed to flow out. Then, without changing the position of the ampoule, it is thoroughly washed with well-boiled distilled water, cooled to a temperature of 35-40°C and taken in such quantity that after cooling the solution to 20°C, only a few drops need to be added to the mark. The titrated alkali solution should be stored in conditions that exclude the possibility of its contact with air.
If there is no fixanal, titrated solutions are prepared from sodium hydroxide (or potassium hydroxide) preparations. The molecular weight of NaOH is 40.01. This number is also its gram equivalent.
To prepare 1 liter 1 i. NaOH solution, you need to take 40 g of chemically pure caustic soda, and to prepare 1 liter 0.1 N. solution - ten times less, i.e. 4 g.
For the convenience of calculating the required amount of starting substances for the preparation of 1 liter of titrated solutions of alkalis of different normalities, we recommend using the data given in Table 31.
Table 31
Initial chemicals, g | Molecular mass | Gram equivalent | Normality of the solution | Substances for installations credits |
||||||
Succinic PLP oxalic acid |
||||||||||
Same |
||||||||||
To prepare 1 liter 0.1 n. sodium hydroxide solution, weigh out a little more than 4 g (4.3-4.5 g) of the drug and dissolve in a small volume of distilled water (about 7 ml).
After settling, the solution is carefully poured (without sediment) into a liter volumetric flask and brought to the mark with distilled freshly boiled water.
The prepared solution is mixed well and placed in a bottle protected from carbon dioxide. After this, the titer is established, i.e. the exact concentration of the solution.
The titer can be set using oxalic or succinic acid. Oxalic acid (C g H 2 0 4 -2H 2 0) is dibasic, and, therefore, its gram equivalent will be equal to half the molecular weight. If the molecular weight of oxalic acid is 126.05 g, then its gram equivalent will be 126.05: 2 = 63.025 g.
The existing oxalic acid should be recrystallized once or twice and only then used to set the titer.
Recrystallization is carried out as follows: take an arbitrary amount of the substance intended for recrystallization, dissolve it by heating, trying to obtain the highest possible concentration of the solution or a saturated solution. If necessary, filter this solution through a hot filter funnel. The filtrate is collected in an Erlenmeyer flask, porcelain cup or glass.
Depending on the nature of crystallization of the substance, the solution, saturated in a hot state, is cooled. To quickly cool the solution during recrystallization, the crystallizer is placed in cold water, snow or ice. With slow cooling, the solution is left to stand at ambient temperature.
If very small crystals fall out, they are dissolved again by heating; the vessel in which the dissolution was carried out is immediately wrapped in several layers with a towel, covered with a watch glass and left to stand completely undisturbed for 12-15 hours.
Then the crystals are separated from the mother liquor, filtered under vacuum (Buchner funnel), thoroughly squeezed, washed and dried.
Preparing 0.1 n. NaOH solution, it is necessary to have a solution of oxalic acid of the same normality; for this, for 1 liter of solution you need to take 63.025: 10 = 6.3025 g. But to set the titer, this amount of oxalic acid solution is too much; It is enough to prepare 100 ml. To do this, about 0.63 g of recrystallized oxalic acid is weighed on an analytical balance, accurate to the fourth decimal place, for example, 0.6223 g. The taken sample of oxalic acid is dissolved in a volumetric flask (per 100 ml). Knowing the mass of the substance taken and the volume of the solution, it is easy to calculate its exact concentration, which in this case is not 0.1 N, but somewhat less.
From the prepared solution, take 20 ml with a pipette, add a few drops of phenolphthalein and titrate with the prepared alkali solution until a faint pink color appears.
Let 22.05 ml of alkali be used for titration. How to determine its titer and normality?
0.6223 g of oxalic acid was taken instead of the theoretically calculated amount of 0.6303 g. Consequently, its normality will not be exactly 0.1
To calculate the normality of alkali, we use the relation VN=ViNt, that is, the product of volume and normality of a known solution is equal to the product of volume and normality for an unknown solution. We get: 20-0.09873 =22.05-a:, from where
To calculate the titer or content of NaOH in 1 ml of solution, the normality should be multiplied by the gram equivalent of the alkali and the resulting product divided by 1000. Then the titer of the alkali will be
But this titer does not correspond to 0.1 n. NaOH solution. To do this, they resort to the coefficient To, i.e., the ratio of the practical titer to the theoretical one. In this case it will be equal
When using succinic acid to establish the titer, its solution is prepared in the same order as oxalic acid, based on the following calculation: the molecular weight of succinic acid (C 4 H 6 0 4) is 118.05 g, but since it is dibasic, its gram equivalent 59.02 g.
To prepare 1 liter of decinormal solution of succinic acid, it must be taken in an amount of 59.02: 10 = 5.902, and for 100 ml of solution -0.59 g.
Setting the titer to 0.1 n. NaOH solution by gravimetric method. To set the titer to 0.1 N. NaOH solution, we take a sample of succinic acid with an accuracy of 0.0001 g (for example, 0.1827 g). Dissolve the sample in distilled water (about 100 ml), then add 3-5 drops of phenolphthalein and titrate with alkali (NaOH). Let's assume that 28 ml of NaOH was used for titration. The calculation of the NaOH titer and the correction to it are carried out as follows: since the gram equivalent of NaOH, equal to 40.01 g, corresponds to the gram equivalent of succinic acid, equal to 59.02 g, then, by making up the proportion, we find out what amount of NaOH contained in weighed amount of succinic acid: 40.01-59.02
We calculate the titer of NaOH, i.e. the content of NaOH in 1 ml of solution. It is equal to: 0.1238: 28 = 0.00442. The correction to the NaOH titer is equal to the ratio of the practical to theoretical titer
Checking the normality of an alkali solution against a titrated acid solution. 20-25 ml of titrated acid solution (HC1 or H2S04) are measured into three conical flasks with a burette and titrated with NaOH solution until the color changes to methyl orange.
Let us assume that for the titration of three samples of 20 ml each, 0.1015 N. HC1 solution consumed an average of 19.50 ml of NaOH solution. The normality of alkali will be
Acid solutions. In most cases, the laboratory has to deal with sulfuric, hydrochloric and nitric acids. They are in the form of concentrated solutions, the percentage of which is determined by density.
For analytical work we use chemically pure acids. To prepare a solution of a particular acid, we usually take the amount of concentrated acids by volume, calculated by density.
For example, you need to prepare 0.1 N. solution H 2 S0 4 . This means that 1 liter of solution should contain 
How much volume do you need to take H 2 S0 4 with a density of 1.84 in order to dilute it to 1 liter to get 0.1 N? solution?
An acid with a density of 1.84 contains 95.6% H 2 S0 4. Therefore, for 1 liter of solution it must be taken in grams:
Expressing the mass in volumetric units, we get
Having measured exactly 2.8 ml of acid from the burette, dilute it to 1 liter in a volumetric flask, then titrate with alkali to check the normality.
For example, during titration it was found that 1 ml of 0.1 N. H 2 S0 4 solution contains not 0.0049 g of H 2 S0 4, but 0.0051 g. To calculate the amount of water that needs to be added to 1 liter of acid, we make up the proportion:
Therefore, 41 ml of water must be added to this solution. But considering that 20 ml of the original solution was taken for titration, which is 0.02, then less water needs to be taken, i.e. 41-(41-0.02) = 41-0.8 = 40.2 ml . This amount of water is added from the burette to the flask with the solution.
The above work is quite painstaking when performed, so you can prepare approximately accurate solutions by introducing a correction factor, which is used in the work for each titration. In this case, we multiply the consumed number of milliliters of solution by a correction factor.
The correction factor is calculated using the formula
Where V - volume of the test solution taken for titration;
k t- correction factor of an alkali solution of known normality, by which the titer of the newly prepared acid solution is established;
Y x is the volume of an alkali solution of known normality used for titration of the test acid.
Table 32
Initial chemicals, ml | Molecular mass | Normality of the solution |
|||||||
{ | |||||||||
H 2 S0 4 (density 1.84) | |||||||||
NS1 (density 1.19) | |||||||||
To facilitate the process of preparing titrated solutions of acids, we offer a table of the amounts of starting substances for preparing 1 liter of solutions of different normalities (Table 32).
It must be kept in mind that when dissolving acids, the acid should be added to the water, and not vice versa.
Dorogov's drug ASD-2 is widely used for the treatment of various diseases in humans and animals. It is intended for both indoor and outdoor use. But most often not in pure form, but in solutions. Today we’ll talk about how to prepare a 1% solution.
How to make a 1% solution of ASD-2 for douching, skin treatment and compresses?
The schemes and methods of using the composition are simple. Scientist A.V. Dorogov has developed several protocols for taking the drug for the treatment of various pathologies. It is according to these schemes that patients are treated. The product is also recommended for external use: lotions, microenemas and vaginal rinsing.
For douching, use a 1% solution. It is very easy to prepare. It is necessary to mix the required number of drops or milliliters of the medicine with boiled, slightly cooled water. The ratio of components is 1:100.
If we take 1 ml of medicine, then it must be mixed with 99 ml of water. How to do it easier and more correctly:
- take 100 ml of boiled water into a measuring cup;
- use a syringe to take 1 ml (cube) of water from the glass, 99 ml remains;
- With another syringe, through a puncture of the rubber stopper, according to the instructions for the drug kit, we collect 1 cube of ASD-2;
- dip the needle of the syringe with the medicine into water;
- carefully squeeze out the drug;
- no additional mixing is required, the medicine itself quickly mixes with water;
- We use the prepared solution immediately, do not store it, otherwise its healing qualities will be lost.
Attention! Do not open the bottle while taking the medicine. When an adaptogen interacts with air, the medicinal properties of the composition are lost, and it simply becomes inactive.
Since the stimulant has a specific, rather unpleasant aroma, it is preferable to mix it with water near an open window and try not to inhale the vapors of the drug.
In what cases should it be used?
External use of an antiseptic stimulant helps cure a wide variety of ailments, including gynecological and skin ones. The medicine has powerful anti-inflammatory, wound-healing, antibacterial and antiseptic effects. Using the solution will help in:
- cure skin diseases: psoriasis, neurodermatitis, trophic ulcers, eczema;
- treatment of skin pathologies of fungal origin;
- accelerating the wound healing process;
- treatment of gynecological ailments: thrush, endometriosis, cervical erosion, uterine fibroids.
Douching with diluted liquid should be done two to three times a day. The duration of the therapeutic course is until complete recovery.
To the question, tell me how to make a 1% solution with copper sulfate and lime (or soda ash) asked by the author Prosphora the best answer is that Bordeaux mixture cannot be prepared with soda ash. Soda ash is sodium carbonate, and you need calcium hydroxide (slaked lime). Otherwise you will get malachite. How to cook - read:
Answer from Mujtahid.[guru]
Answer from Sedge[guru]
Answer from Elena Akentyeva[guru]
Answer from chevron[guru]
to prepare a 1% solution for copper, you need to dilute 100 g of vitriol in 10 liters of water and neutralize 100 g of slaked lime ==>>
Answer from Natali Natali[guru]
Answer from Zhanna S[guru]
1% Bordeaux mixture.
and also dilute 5 liters of water.
(by green leaves).
1-% BURGUNDY liquid
Add 50 g household goods. soap
COPPER_SOAP SOLUTION.
+soap in the same quantity.
Success in the fight against disease)
Answer from Mujtahid.[guru]
It is better to buy ready-made Bordeaux mixture.
Answer from Galina Ruskova (churkina) GALJ[guru]
TREATING PLANTS AGAINST PHYTOPHORA AND OTHER DISEASES
Answer from Elena Akentyeva[guru]
Do not suffer with Bordeaux mixture, it is a very inconvenient preparation in terms of preparation (does not mix well) and processing (it clogs the sprayer). Buy Ordan or Abiga-Pik, wonderful fungicides, no hassle.
Answer from Kostenko Sergey[guru]
to prepare a 1% solution for copper, you need to dilute 100 g of vitriol in 10 liters of water and neutralize 100 g of slaked lime ==>> 10 liters of Bordeaux mixture with a copper concentration of 1%
Answer from Natali Natali[guru]
Dilute 100g of copper sulfate in 5 liters of warm water and separately dilute 100g of lime in 5 liters of water. Then POUR the vitriol solution into the lime solution - NOT vice versa and get a 1% solution of Bordeaux mixture. In other words: for 10 liters of water - 100 g of vitriol and lime
Answer from Zhanna S[guru]
Bordeaux mixture is prepared on the basis of copper sulfate and lime.
BURGUNDY - made from copper sulfate and soda (you can also use baking soda) + soap.
1% Bordeaux mixture.
100 g of quicklime and slaked in a small amount of water, diluted with water to 5 liters, lime milk is obtained.
In another container (non-metallic)
dissolve 100 g of copper sulfate in hot water
and also dilute 5 liters of water.
Pour a solution of copper sulfate into the milk of lime and stir well.
Copper into lime, not the other way around!
You can dissolve 100 g of copper sulfate in 1 liter of hot water
and pour, gradually stirring, into 9 liters of lime milk.
But mix both concentrated solutions,
and then diluting it with water to 10 liters is unacceptable.
The result is a mixture of poor quality.
A properly prepared liquid has a turquoise, sky blue color and a neutral or slightly alkaline reaction.
Acidity is checked with litmus paper,
which turns blue.
You can put any clean (not rusty) piece of iron into the solution.
In an acidic environment, copper actively deposits on iron.
The acidic mixture will burn the leaves.
It is neutralized by adding lime milk.
A 1% mixture is used on vegetative plants.
(by green leaves).
1-% BURGUNDY liquid
100 g of copper sulfate and 100 g of soda ash per 10 liters of water. Dissolve separately
drain together, check acidity,
that is, it is prepared in the same way as the Bordeaux mixture, only the lime is replaced with soda.
Add 50 g household goods. soap
COPPER_SOAP SOLUTION.
10 g of copper sulfate is dissolved in 0.5 liters of hot water.
Separately, 100 g of soap are diluted in 10 liters of water (preferably warm).
The copper sulfate solution is poured into the soap solution in a thin stream with constant stirring.
The drug is prepared before spraying.
The copper-soap preparation (emulsion) can be prepared in higher concentrations
(20 g copper sulfate and 200 g soap
or 30 g of vitriol and 300 g of soap per 10 liters of water).
A properly prepared emulsion should have a greenish color and not form flakes.
To avoid curdling of the drug when preparing it in hard water,
the amount of copper sulfate should be reduced
or add 0.5% (50 g per 10 liters of water) soda ash (linen) to the water.
Can be used together with karbofos (20 g per 10 l of emulsion)
for simultaneous control of aphids and spider mites.
SODA ASH, or laundry soda (sodium carbonate) is a white crystalline powder, soluble in water.
Used to control powdery mildew,
at a concentration of 0.5% (50 g per 10 l of water).
+soap in the same quantity.
To prepare a working solution, dilute soap in soft water and add soda, previously dissolved in a small amount of water.
Success in the fight against disease)
To prepare solutions of molar and normal concentrations, a sample of the substance is weighed on an analytical balance, and the solutions are prepared in a volumetric flask. When preparing acid solutions, the required volume of concentrated acid solution is measured with a burette with a glass stopcock.
The weight of the solute is calculated to the fourth decimal place, and the molecular weights are taken with the accuracy with which they are given in the reference tables. The volume of concentrated acid is calculated accurate to the second decimal place.
Example 1. How many grams of barium chloride are needed to prepare 2 liters of 0.2 M solution?
Solution. The molecular weight of barium chloride is 208.27. Therefore. 1 liter of 0.2 M solution should contain 208.27-0.2= = 41.654 g BaCl 2 . To prepare 2 liters you will need 41.654-2 = 83.308 g of BaCl 2.
Example 2. How many grams of anhydrous soda Na 2 C0 3 are required to prepare 500 ml of 0.1 N. solution?
Solution. The molecular weight of soda is 106.004; equivalent unit mass 5 N a 2 C0 3 =M: 2 = 53.002; 0.1 eq. = 5.3002 g
1000 ml 0.1 n. solution contain 5.3002 g Na 2 C0 3
500 »» » » » X
» Na 2 C0 3
5,3002-500
x=—— Gooo—- = 2-6501 g Na 2 C0 3.
Example 3. How much concentrated sulfuric acid (96%: d=l.84) is required to prepare 2 liters of 0.05 N. sulfuric acid solution?
Solution. The molecular weight of sulfuric acid is 98.08. Equivalent mass of sulfuric acid 3h 2 so 4 = M: 2 = 98.08: 2 = 49.04 g. Mass 0.05 eq. = 49.04-0.05 = 2.452 g.
Let's find how much H 2 S0 4 should be contained in 2 liters of 0.05 n. solution:
1 l-2.452 g H 2 S0 4
2"- X » H 2 S0 4
X = 2.452-2 = 4.904 g H 2 S0 4.
To determine how much 96.% H 2 S0 4 solution needs to be taken for this, let’s make a proportion:
\ in 100 g of conc. H 2 S0 4 -96 g H 2 S0 4
U» » H 2 S0 4 -4.904 g H 2 S0 4
4,904-100
U=——— §6—— = 5.11 g H 2 S0 4 .
We recalculate this quantity to volume: ,. R 5,11
K = 7 = TJ = 2 ‘ 77 ml -
Thus, to prepare 2 liters of 0.05 N. solution you need to take 2.77 ml of concentrated sulfuric acid.
Example 4. Calculate the titer of a NaOH solution if it is known that its exact concentration is 0.0520 N.
Solution. Let us recall that the titer is the content in 1 ml of a solution of a substance in grams. Equivalent mass of NaOH = 40 01 g Let's find how many grams of NaOH are contained in 1 liter of this solution:
40.01-0.0520 = 2.0805 g.
1 liter of solution: -n=- =0.00208 g/ml. You can also use the formula:
9 N
Where T- titer, g/ml; E- equivalent mass; N- normality of the solution.
Then the titer of this solution is:
f 40,01 0,0520
“NaOH =——— jooo—— 0.00208 g/ml.
„ “Rie P 5 - Calculate the normal concentration of the HN0 3 solution if it is known that the titer of this solution is 0.0065. To calculate, we use the formula:
T ■ 1000 63,05
5hno 3 = j- = 63.05.
The normal concentration of nitric acid solution is:
- V = 63.05 = 0.1030 n.
Example 6. What is the normal concentration of a solution if it is known that 200 ml of this solution contains 2.6501 g of Na 2 C0 3
Solution.
As was calculated in example 2, Zma 2 co(=53.002.
Let's find how many equivalents are 2.6501 g of Na 2 C0 3: G
2.6501: 53.002 = 0.05 eq. /
In order to calculate the normal concentration of a solution, we create a proportion:
1000 » » X "
1000-0,05
x =
—————— =0.25 eq.
1 liter of this solution will contain 0.25 equivalents, i.e. the solution will be 0.25 N.
For this calculation you can use the formula:
R- 1000
Where R - amount of substance in grams; E - equivalent mass of the substance; V - volume of solution in milliliters.
Zia 2 with 3 = 53.002, then the normal concentration of this solution
2.6501-10С0 N = 53.002-200
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